Answer
a) The force has a magnitude of $9.84N$ and is directed in the $-x$ direction.
b) $\omega=10.5rad/s$
c) $v_{max}=1.25m/s$
d) $a_{max}=13.12m/s^2$
Work Step by Step
a) We know the spring constant $k=82N/m$ and the displacement $x=+0.12m$. The force the spring exerts on the block (restoring force) just before the block is released is $$F_x=-kx=-9.84N$$
So the force has a magnitude of $9.84N$. The negative sign indicates that the force is directed in the $-x$ direction.
b) The angular frequency of the motion is $$\omega=\sqrt{\frac{k}{m}}$$
The block's mass $m=0.75kg$, so $$\omega=10.5rad/s$$
c) The displacement the block is given before it is released is the amplitude of the motion, so $A=0.12m$
The maximum speed, therefore, is $$v_{max}=A\omega=1.25m/s$$
d) The maximum acceleration is $$a_{max}=A\omega^2=13.12m/s^2$$
